https://i.imgur.com/CIiyB2D.jpg
a, \(n_{Mg}=\frac{9,6}{24}=0,4\left(mol\right)\)
\(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
\(2Mg+O_2\rightarrow2MgO\)
Lập tỉ lệ : \(\frac{0,4}{2}< \frac{0,5}{1}\)
\(\rightarrow\) Mg hết , O2 còn dư
b, Theo phương trình :
\(n_{O2_{pu}}=\frac{1}{2}n_{Mg}=\frac{1}{2}.0,4=0,2\left(mol\right)\)
\(n_{O2_{du}}=0,5-0,2=0,3\left(mol\right)\)
\(m_{O2_{du}}=0,3.32=6,4\left(g\right)\)
c, \(n_{MgO}=n_{Mg}=0,4\)
\(\rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
